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  "body": "\n## Problem\n\nFix a field $F$.\nConsider the real numbers $\\tau$ such that an algebraic algorithm multiplies two $n \\times n$ matrices over $F$ with $O(n^\\tau)$ operations in $F$ (see [[big O notation|0TGV80E6A4]]).\nThe *matrix multiplication exponent* $\\omega$ is the infimum of these $\\tau$.\nDetermine the value of $\\omega$; in particular, decide whether $\\omega = 2$.\n\n## Remarks\n\nThe bound $\\omega \\ge 2$ holds because the output has $n^2$ entries, and the schoolbook [[matrix multiplication|032Q9FB8MT]] gives $\\omega \\le 3$.\nThe [[Strassen algorithm|3CVWXDS4JQ]] gave $\\omega \\le \\log_2 7 \\approx 2.807$ in 1969.\nCoppersmith and Winograd obtained $\\omega \u003c 2.376$ in 1990.\nRefinements of their method by several groups since 2010 have brought the upper bound below $2.372$.\nThe problem is open: no lower bound better than $2$ is known.\nStrassen showed that $\\omega$ can be read off from the [[tensor rank|W84NTF77M0]] of the matrix multiplication tensor $\\langle n, n, n \\rangle$.\nIt is the infimum of the $\\tau$ such that this rank is $O(n^\\tau)$.\nMany other problems of linear algebra, such as inverting a matrix or computing a determinant, can be solved with $O(n^{\\omega + \\varepsilon})$ operations for every $\\varepsilon \u003e 0$.\n\n## Explanations\n\n### L1 human\n\nMultiplying two $n \\times n$ matrices the usual way needs about $n^3$ multiplications.\nThe exponent $\\omega$ measures how much better one can do in the best case, as a power of $n$.\nStrassen showed that the power $3$ is not optimal, and later work has pushed it down to about $2.37$.\nNobody knows whether one can get as low as $2$, which would mean that multiplying matrices costs hardly more than reading them.\n",
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